Question

A balloon is ascending vertically with acceleration of 0.2$m{{s}^{-2}}$. Two stones are dropped from it at an interval of 2 sec. The distance (in m) between them 1.5 sec after the second stone is released is approximately 10x. the value of x is (use $g=10m{{s}^{-2}}$)

Answer 1

Hint: In the above question it is given that the two stones are released from a balloon which is accelerating upwards. As soon as the stones are released from the balloon, considering the balloon to be at rest, the initial velocity of the stones dropped will also be zero. But the upwards acceleration of the balloon has to be given to the stone in the downward direction as we have taken the balloon to be at rest. Hence using Newton’s second kinematic equation, we can determine the distance of separation between the stones 1.5 sec after the release of the second stone.

Complete step-by-step solution:
Let us say we have a body having initial velocity ‘u’. If the acceleration of the body is ‘a’, then at time t the distance covered ‘s’ by the body is given by,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
In the above question it is given that the two stones are dropped at an interval of 2sec and we are asked to determine the distance between them after an interval of 1.5sec from the release of the second stone. Hence we can imply that the total time of flight of the first stone is 3.5sec whereas that of the second stone is 1.5sec.
Now to solve the above question let us say we analyze the above situation with respect to the balloon. Let us say the balloon is at rest. Hence we can imply that the initial velocity of the stones dropped is also zero. For a observer on the balloon he will experience that the stones dropped are going down due to acceleration due to gravity ‘g’ as well it is moving with an acceleration of 0.2$m{{s}^{-2}}$. hence the net acceleration ‘a’ of the stones dropped with respect to the balloon is,
$\begin{align}
  & a=g+0.2m{{s}^{-2}} \\
 & \Rightarrow a=10m{{s}^{-2}}+0.2m{{s}^{-2}} \\
 & \therefore a=10.2m{{s}^{-2}} \\
\end{align}$
From Newton’s second kinematic equation, the total distance ${{S}_{1}}$covered by the stone in the interval of 3.5sec is,
\[\begin{align}
  & {{S}_{1}}=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & \because a=10.2m{{s}^{-2}},\text{ }u=0m{{s}^{-1}} \\
 & \Rightarrow {{S}_{1}}=(0m{{s}^{-1}})3.5s+\dfrac{1}{2}\left( 10.2m{{s}^{-2}} \right){{(3.5s)}^{2}} \\
 & \Rightarrow {{S}_{1}}=\dfrac{1}{2}\times 10.2m\times 12.25 \\
 & \therefore {{S}_{1}}=62.475m \\
\end{align}\]
Similarly the distance covered ${{S}_{2}}$by the second stone is in the interval of 1.5sec is,
\[\begin{align}
  & {{S}_{2}}=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & \because a=10.2m{{s}^{-2}},\text{ }u=0m{{s}^{-1}} \\
 & \Rightarrow {{S}_{2}}=(0m{{s}^{-1}})1.5s+\dfrac{1}{2}\left( 10.2m{{s}^{-2}} \right){{(1.5s)}^{2}} \\
 & \Rightarrow {{S}_{2}}=\dfrac{1}{2}\times 10.2m\times 2.25 \\
 & \therefore {{S}_{2}}=11..475m \\
\end{align}\]
Therefore the distance ‘d’ between the two stones after an interval of 1.5s from the release of the second stone is,
$\begin{align}
  & d={{S}_{1}}-{{S}_{2}} \\
 & \Rightarrow d=62.475m-11.475m \\
 & \therefore d=51m \\
\end{align}$

Note:It is to be noted that the above problem can also be solved with respect to ground. To understand the above problem consider yourself to be on the balloon and at every instant of time just consider yourself to be at rest. The understanding of the pseudo force will be more effective in such cases.